Bunuel wrote:

A “descending number” is a three-digit number, such that the units digit is less than the tens digits and the tens digit is less than the hundreds digit. What is the probability that a three-digit number chosen at random is a “descending number”?

(A) 3/25

(B) 2/9

(C) 2/15

(D) 1/9

(E) 1/10

Kudos for correct solution.

We see that the hundreds digit can be any digit from 2 to 9.

If we start with the hundreds digits as 2, then the tens digit must be 1, and the units digit must be 0. We have 1 descending number when the hundreds digit is 2.

If the hundreds digits is 3, then if the tens digit is 2, the units digit can be1 or 0; if the tens digit is 1, the units digit must be 0. We have 3 descending numbers when the hundreds digit is 3.

Now, if the hundreds digit is 4, then if the tens digit is 3, the units digit can be 2, 1, or 0; if the tens digit is 2, the units digit can be1 or 0; if the tens digit is 1, the units digit must be 0. We have 6 descending numbers when the hundreds digit is 4.

From this point, we can see a pattern:

hundreds digit = 2 → number of descending numbers = 1

hundreds digit = 3 → number of descending numbers = 1 + 2 = 3

hundreds digit = 4 → number of descending numbers = 1 + 2 + 3 = 6

So the number of descending numbers when the hundreds digit is 5 is 10, 6 is 15, 7 is 21, 8 is 28, and 9 is 36. Therefore, there are a total of 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120 three-digit descending numbers.

Since there are 999 - 100 + 1 = 900 three-digit numbers, the probability of picking a descending number is 120/900 = 12/90 = 2/15.

Answer: C

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